Why I can't exit code in python?

Question

I've tried use exit(), exit(0), sys.exit()(I've import sys) and quit(), but none of them can help me to exit code.

Here is my code when using exit():

from pystray import MenuItem as item
import pystray
from PIL import Image


while True:
    def show():
        exit()
    image = Image.open("TrayIcon.jpg")
    menu = (item('exit', show),)
    icon = pystray.Icon("name", image, "title", menu)
    icon.run()
    print('running...')

This is the error when using exit()

An error occurred when calling message handler
Traceback (most recent call last):
  File "D:\py3.7\lib\site-packages\pystray\_win32.py", line 402, in _dispatcher
    uMsg, lambda w, l: 0)(wParam, lParam) or 0)
  File "D:\py3.7\lib\site-packages\pystray\_win32.py", line 213, in _on_notify
    descriptors[index - 1](self)
  File "D:\py3.7\lib\site-packages\pystray\_base.py", line 324, in inner
    callback(self)
  File "D:\py3.7\lib\site-packages\pystray\_base.py", line 449, in __call__
    return self._action(icon, self)
  File "D:\py3.7\lib\site-packages\pystray\_base.py", line 544, in wrapper0
    return action()
  File "C:/Users/admin/AppData/Roaming/JetBrains/PyCharmCE2022.1/scratches/scratch.py", line 9, in show
    exit()
  File "D:\py3.7\lib\_sitebuiltins.py", line 26, in __call__
    raise SystemExit(code)
SystemExit: None

i aslo have tried to turn

from pystray import MenuItem as item
import pystray
from PIL import Image


def show():
    icon.stop()


image = Image.open("TrayIcon.jpg")
menu = (item('exit', show),)
icon = pystray.Icon("name", image, "title", menu)
icon.run()

while True:
    print('running...')

At this time, i can't running... is not show when the icon is in the tray, i must exit it to show.

Other errors are quiet similar with the error when using exit()

The **purpose of** `exit`, and all its friends, is to raise that exact exception. Normally, it is handled by the Python interpreter itself, but here it has been caught and displayed by `pystray`. — Karl Knechtel, Aug 15 '22 at 04:32
@Ashkan Goleh Pour I've call it in `menu = (item('exit', show),)` — Danhui Xu, Aug 15 '22 at 04:35
in the `while` loop your function `show()` gets decleared which contains `exit()` but it is never called any where. — Muhammad Zakaria, Aug 15 '22 at 04:36
Never use `exit` or `quit` in your program. To cite the [documentation of `exit` and `quit`](https://docs.python.org/3/library/constants.html#exit): "They are useful for the interactive interpreter shell and should not be used in programs." — Matthias, Aug 15 '22 at 07:25

Mohamad Ghaith Alzin · Answer 1 · 2022-08-15T05:05:43.210

0

I've given this a test now and you don't need to call exit(). Instead, call icon.stop()

Also, no need for the infinite loop!

from pystray import MenuItem as item
import pystray
from PIL import Image

def show():
    icon.stop()
    
image = Image.open("TrayIcon.jpg")
menu = (item('exit', show),)
icon = pystray.Icon("name", image, "title", menu)
icon.run()

Note: the following link would be useful for you as well

https://github.com/moses-palmer/pystray/issues/17

Pystray systray icon

edited Aug 15 '22 at 05:05

answered Aug 15 '22 at 04:53

Mohamad Ghaith Alzin

869
1
7
18

Sorry, maybe my question is not clear, i must use `while True` in the program. and when i add `icon.stop()` in `def show()`, it tells me `NameError: name 'icon' is not defined` – Danhui Xu Aug 15 '22 at 04:59
What is your reason behind calling an infinite loop!? – Mohamad Ghaith Alzin Aug 15 '22 at 05:00
Because i'm making a programme amd this only a small part of it. In that programme it need to do a lot of things in the loop. That's why i need a loop – Danhui Xu Aug 15 '22 at 05:11
I can't understand your point. Anyway, please edit your question and say exactly what you want to do and why you want your loop each time to create an icon!? – Mohamad Ghaith Alzin Aug 15 '22 at 05:17
When I turn these out of loop, the programm will stop as soon as i start it. – Danhui Xu Aug 15 '22 at 05:28
On my side, when I run the program just ends when I press on exit and then the icon gets removed!! – Mohamad Ghaith Alzin Aug 15 '22 at 05:50
Well, i'm to output `running...` before i exit the programm in the tray, at this time, i must exit the tray icon, then the things in the `While True:` can be output. – Danhui Xu Aug 15 '22 at 08:33
@DanhuiXu Do mean you want to print `running...` one time when the exit is pressed? if so, just put the print it in the show() function! – Mohamad Ghaith Alzin Aug 15 '22 at 08:51
Sorry,l mean print `running...` before exit. – Danhui Xu Aug 15 '22 at 10:52
Ok ,just do it before `icon.stop()`!! – Mohamad Ghaith Alzin Aug 15 '22 at 11:56
Sorry,I don't understand.Could you said more clearly,Thanks – Danhui Xu Aug 16 '22 at 11:29
@DanhuiXu I mean put the `print` in the `show()` function before the `stop` statement – Mohamad Ghaith Alzin Aug 16 '22 at 12:41

Gajera Yugantkumar Rajanikumar · Answer 2 · 2022-08-15T05:05:07.690

-1

To exit the shell and return to the system prompt, type exit() or Ctrl-D.

edited Aug 15 '22 at 05:05

answered Aug 15 '22 at 05:04

Gajera Yugantkumar Rajanikumar

1
2

Well, I'm not working in a shell I'm working in the IDE and doing a programe. – Danhui Xu Aug 15 '22 at 05:07
use the same code. – Gajera Yugantkumar Rajanikumar Aug 15 '22 at 05:09
Maybe the question is not clear, In my question i've said that `exit()` is not working in my code. – Danhui Xu Aug 15 '22 at 05:13
This is a comment! not an answer. – Mohamad Ghaith Alzin Aug 15 '22 at 07:25
As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Aug 18 '22 at 09:27

Why I can't exit code in python?

2 Answers2